First, a terminological nitpick: the Schur-Weyl duality deals with the *unitary* group $U(N)$ (in the compact formulation) or the general linear group $GL(N)$ (in the algebraic group version) acting on $\mathcal{H}^{\otimes m}$, where $\mathcal{H}=\mathbb{C}^N.$ The duality states that the isotypic components under the $U(N)$ action are irreducible $S_m$-modules. Explicitly,

$$\mathcal{H}^{\otimes m}\simeq \bigoplus_{\lambda}\rho_{U(N)}^\lambda\otimes\rho_{S_m}^{\lambda},$$

where $\lambda$ runs over Young diagrams with $m$ boxes and at most $\operatorname{min}(N,m)$ rows. Observe that for $m\geq 3$ and $N\geq 2$ this decomposition is *not* multiplicity-free as a $U(N)$-module. However, for $N\geq 2$ the restriction to $SU(N)$ does not introduce new multiplicities: for distinct $\lambda,\mu$ as above, the representations $\rho_{SU(N)}^\lambda$ and $\rho_{SU(N)}^\mu$ are non-isomorphic .

Now for the present question. Consider the $K$-equivariant isomorphism $$\mathcal{H}^{\otimes m}\simeq \mathcal{H}_1^{\otimes m} \otimes\ldots\otimes\mathcal{H}_s^{\otimes m}.$$ Then each factor $SU(N_i)$ of $K$ acts on its own $m$th tensor power space $\mathcal{H}_i^{\otimes m}.$ It is a standard fact in representation theory that the irreducible representations of $K=SU(N_1)\times\ldots\times SU(N_s)$ have the form $V_1\otimes\ldots\otimes V_s,$ where $V_i$ is an irreducible representation of $SU(N_i)$ determined uniquely up to isomorphism. Hence the question is reduced to the case $s=1.$ Explicitly,
$$\mathcal{H}^{\otimes m}\simeq \bigoplus_{\lambda_1,\ldots,\lambda_s}\rho_{SU(N_1)}^{\lambda_1}\otimes\ldots\otimes \rho_{SU(N_s)}^{\lambda_s}\otimes\rho_{S_m}^{\lambda_1}\otimes\ldots\otimes\rho_{S_m}^{\lambda_s}.$$

Different $\lambda_i$'s with $m$ boxes are independently chosen, each subject to its own restriction on the number of rows. This is not multiplicity-free unless $m\leq 2$ or all $N_i$ are equal to 1.